My Math Forum Projection Transformation on x Axis Parallel to y=2x

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 October 14th, 2015, 09:57 PM #1 Newbie   Joined: Oct 2015 From: England Posts: 2 Thanks: 0 Projection Transformation on x Axis Parallel to y=2x Hello, Let $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the projection transformation on x axis in parallel to $y=2x$. Find the representing matrix of $T^*$ for the standard basis $B=\left(\left \{1,0 \right \},\left\{0,1\right\}\right)$. I have a solution which I don't understand. A basis for $y=2x$ could be $\left \{(1,2) \right\}$. A basis for x axis could be $\left \{(1,0)\right\}$. Therefore A basis for the whole vector would be $$C=\left\{(1,0)\right\},\left\{(1,2)\right\}$$. I understand why this basis is correct. However: $e_1=(1,0)=1v_1+0v_2 \Rightarrow [e_1]_B=\left (\begin {matrix} 1\\ 0 \end {matrix}\right)$ - this is quite clear. $e_2=(0,1)=-\frac{1}{2}v_1+\frac{1}{2}v_2\Rightarrow [e_2]_B=\left (\begin {matrix} -\frac{1}{2} \\ \frac{1}{2} \end {matrix}\right)$ - this is the part I don't understand. Why use $-\frac{1}{2}$ to multiply $v_1$? What am I missing here? Next, since the proejction transformation is $T(x,y)=(x,0)$ it is easy enough to continue. But there's that part I don't understand. Why is the scalar before the vector $v_1$has to be$-\frac{1} {2}$ Thanks, Alan
 October 15th, 2015, 09:20 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Given unit vector u. Projection of any vector x onto u in matrix form is given by Tx where $\displaystyle T_{ij} =u_{i}u_{j}$ Change of basis: e1'=P11e1+P21e2 e2'=P12e1+P22e2 Relative to new basis: $\displaystyle T'=P^{-1}TP$ Reference: https://www.khanacademy.org/math/lin...ix-vector-prod Linear Transformations. EDIT: I think I see now what you are asking: Find the projection of x onto x axis in direction of line y=2x. Have to think about how to put that in matrix form. Last edited by zylo; October 15th, 2015 at 09:28 AM.
 October 15th, 2015, 10:08 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Projection of x in direction u on x axis. ae1+bu=x and solve for a a=x1-x2u1/u2 In matrix form: Tx= $\displaystyle \begin{bmatrix} 1 &-u1/u2 \\ 0& 0 \end{bmatrix}\begin{bmatrix} x1\\ x2 \end{bmatrix}=\begin{bmatrix} x1-x2u1/u2\\ 0 \end{bmatrix}$ Last edited by zylo; October 15th, 2015 at 10:20 AM.
 October 19th, 2015, 09:47 AM #4 Newbie   Joined: Oct 2015 From: England Posts: 2 Thanks: 0 Thank you both!
October 20th, 2015, 08:33 AM   #5
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 Originally Posted by Nerazzurri10 Thank you!
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 October 31st, 2015, 07:17 AM #6 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Systematic Solution to OP based on first principles. T:Rn->Rm is a linear transformation if T(au+bv)=aTu+bTv Matrix equivalent of T:R2->R2 T(x1e1+x2e2)=x1Te1+x2Te2=x1(T11e1+T21e2)+x2(T12e1+ T22e2) Tij are the matrix components of T Tx=[T]x Example: Project x onto line w in direction v u: unit vector in direction of x v: unit vector in direction of projection w: unit vector along line which x is to be projected onto, au+bv=cw (draw a sketch, all vectors from origin) Te1=cw when au=e1 e1+bv=cw 1+bv1=cw1 bv2=cw2, and solve: c=v2/(w1v2-w2v1) Te2=c'w when a'u=e2 e2+b'v=c'w b'v1=c'e1 1+b'v2=c'w2, and solve c'=-v1/(w1v2-w2v1) Te1=cw=c(w1e1+w2e2)=T11e1+T21e2 Te2=c'w=c'(w1e1+w2e2)=T12e1+T22e2 T11=cw1, T21=cw2, T12=c'w1, T22=c'w2 EDIT: That's the general case. For OP line to be projected onto is x axis so w1=1, w2=0, and also v1=1/sqrt(5), v2=2/sqrt(5) Last edited by zylo; October 31st, 2015 at 07:30 AM.

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